Definitions Flashcards

1
Q

Scalar quantity

A

A quantity that only has magnitude and no direction like mass

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2
Q

Vector quantity

A

A quantity that has both magnitude and direction

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3
Q

Examples of vector quantities

A

Displacement
Velocity
Force
Weight
Acceleration

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4
Q

Examples of scalar quantities

A

Mass
Energy
Time
Distance
Speed

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5
Q

Displacement is defined as

A

A vector quantity measuring a change in position. How much an object has changed and what direction the motion takes place to be able to locate the new position of the object

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6
Q

Displacement is measured in

A

M
Mm to m÷1000
Cm to m÷100

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7
Q

An angle can move

A

Up down, left right, angle initial direction in angle, compass points or bearing.

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8
Q

North of East

A

You are heading to a north side from an east point.

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9
Q

Direction help

A

“North of East” and “East of North” may seem similar, but they have distinct meanings:

“North of East” (NE):

  • Start facing East
  • Turn slightly North (less than 45°)

“East of North” (EN):

  • Start facing North
  • Turn slightly East (less than 45°)

Key difference:

  • “North of East” leans more towards East
  • “East of North” leans more towards North

Visualize:

  • Imagine a compass rose
  • Draw a line from North to East
  • “North of East” falls slightly below this line
  • “East of North” falls slightly above this line

Mnemonic:

  • “North of East” = Need Extra Energy
  • “East of North” = Enter New Nation

Practice:

  • Use online compass simulators
  • Practice navigation with maps
  • Familiarize yourself with directional notation

Now, you can confidently distinguish between “North of East” and “East of North”!

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10
Q

Pythagoras theorem and displacement

A

Pythagoras’ theorem helps find displacement when you know the horizontal and vertical components of the motion.

Pythagoras’ Theorem:

a² + b² = c²

where:

  • a = horizontal displacement (or distance)
  • b = vertical displacement (or distance)
  • c = resultant displacement (hypotenuse)

Finding Displacement:

  1. Break down the motion into horizontal and vertical components.
  2. Calculate the horizontal (a) and vertical (b) displacements.
  3. Apply Pythagoras’ theorem to find the resultant displacement (c).

Example:

An object moves:

  • 3 meters east (horizontal)
  • 4 meters north (vertical)

Find the resultant displacement.

Solution:

a = 3 m (east)
b = 4 m (north)
c = ? (resultant displacement)

Using Pythagoras’ theorem:

c² = a² + b²
c² = 3² + 4²
c² = 9 + 16
c² = 25
c = √25
c = 5 m

Resultant Displacement: 5 meters

Direction: Use trigonometry (tan, cos, or sin) to find the direction.

tan(θ) = b/a
tan(θ) = 4/3
θ ≈ 53.13° north of east

Resultant Vector: 5 m, 53.13° north of east

Pythagoras’ theorem is essential in physics, engineering, and navigation for calculating displacements, velocities, and forces in two-dimensional space!

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11
Q

What about direction and trig ratios

A

Trigonometric ratios (trig ratios) help find the direction of the resultant vector.

Trigonometric Ratios:

  1. Sine (sin): opposite side / hypotenuse
  2. Cosine (cos): adjacent side / hypotenuse
  3. Tangent (tan): opposite side / adjacent side

Finding Direction:

  1. Identify the horizontal (adjacent) and vertical (opposite) components.
  2. Choose the appropriate trig ratio based on the given information.

Examples:

1. Finding angle (θ) using tangent:

Horizontal displacement (a) = 3 m
Vertical displacement (b) = 4 m

tan(θ) = b / a
tan(θ) = 4 / 3
θ = arctan(4/3)
θ ≈ 53.13° north of east

2. Finding angle (θ) using sine and cosine:

Resultant displacement (c) = 5 m
Vertical displacement (b) = 4 m

sin(θ) = b / c
sin(θ) = 4 / 5
θ = arcsin(4/5)
θ ≈ 53.13° north of east

cos(θ) = a / c
cos(θ) = 3 / 5
θ = arccos(3/5)
θ ≈ 53.13° north of east

Important:

  • Use the correct trig ratio based on the given information.
  • Consider the quadrant in which the angle lies.

Quadrants:

  • I (0-90°): north-east
  • II (90-180°): north-west
  • III (180-270°): south-west
  • IV (270-360°): south-east

Trigonometric ratios are essential in physics, engineering, and navigation for calculating angles and directions in two-dimensional space!

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12
Q

In a graph of motion a straight line tells us that

A

The position of an object changes by a constant amount in each second

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13
Q

The slope of a position time graph

A

Slope= change in position (final position - initial position =0)
So delta x over delta t equals velocity

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14
Q

A straight line graph

A

Has constant speed bc the velocity is also constant. If you cycle faster then your velocity increases.

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15
Q

Negative graph of motion

A

Slope goes down and negative gradient.

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16
Q

Position is in

A

Meters

17
Q

Velocity is in

A

M/s - 1

18
Q

Acceleration

A

Rate of change of velocity with time. A= delta x over delta t

19
Q

Vf

A

Final velocity of an object. Meters per second to the power of negative 1.

20
Q

Vi

A

Initial velocity. Meters epr second to the power of negative 1

21
Q

A

A

Acceleration. Meters per second to the power of negative 2

22
Q

Delta t

A

Time taken in seconds

23
Q

First equation

A

A = vi- vf over delta t.
Uniformly accelerated motion

24
Q

Equation 2 : displacement

A

Delta x= 1/2 ( initial velocity + final velocity) times delta x

25
Q

Equation 3 :find final velocity

A

Vf= initial velocity + acceleration times delta t

26
Q

Average velocity

A

Sum of initial and final divided by 2